$\ce{Pb^4+}$ is reduced by $\ce{I-},$ so $\ce{PbI4}$ cannot exist with $\ce{Pb}$ in $+4$ oxidation state. But if I replace three $\ce{I-}$ with $\ce{I3-},$ then we can have $\ce{Pb^2+}$ similar to $\ce{TlI3}$ which exists with $\ce{Tl+}.$ Why does this not happen?
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4$\begingroup$ $\ce{Pb^2+}$ prefers $\ce{[PbI4]^2-}$. $\endgroup$– PoutnikMay 26, 2021 at 7:21
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1$\begingroup$ No idea, I cannot confirm nor reject it. $\endgroup$– PoutnikMay 26, 2021 at 11:50
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2$\begingroup$ Partly because I have never even heard of a triiodide salt with a 2+ cation. Such compounds would likely decompose to give the normal iodide salt plus iodine. Even with a heavy alkali metal such as caesium, the triiodide ion is in effect partly decomposed, with one bond becoming longer than the other as if you had a complex of $\ce{I2}$ with a normal (mono)iodide ion. $\endgroup$– Oscar LanziMay 26, 2021 at 15:01
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3$\begingroup$ Does this answer your question? Why triiodide ion does not form ionic bond with cation having +2 charge? $\endgroup$– Nilay GhoshMay 27, 2021 at 3:58
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1$\begingroup$ @nilayghosh Sure, depending on $\ce{[I-]}$ :-) One cannot get one without another. $\endgroup$– PoutnikMay 27, 2021 at 4:32
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